Generators for the elliptic curve y2 = x3− nx Tome 23, no2 (2011), p. 403-416.
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Generators for the elliptic curve y
2= x
3− nx
par Yasutsugu FUJITA et Nobuhiro TERAI
Résumé. Soit E la courbe elliptique définie par y2= x3− nx où
n est un entier strictement positif. En 2007, Duquesne a démontré
que, pour k entier, si n = (2k2− 2k + 1)(18k2+ 30k + 17) est
sans facteur carré, alors deux points rationnels spécifiques peuvent toujours se compléter en un système de générateurs du groupe de Mordell-Weil associé à E. Dans ce papier, nous généralisons ce résultat en le montrant pour des entiers n = n(k, l) pour une infinité de formes binaires n(k, l) ∈ Z[k, l].
Abstract. Let E be an elliptic curve given by y2 = x3− nx with a positive integer n. Duquesne in 2007 showed that if n = (2k2− 2k + 1)(18k2+ 30k + 17) is square-free with an integer k,
then certain two rational points of infinite order can always be in a system of generators for the Mordell-Weil group of E. In this paper, we generalize this result and show that the same is true for infinitely many binary forms n = n(k, l) in Z[k, l].
1. Introduction
Let E be an elliptic curve over the rationals Q defined by
E : y2 = x3− nx
with a positive integer n. Mordell’s theorem asserts that the group E(Q) of rational points on E is finitely generated. It is easy to check that the torsion subgroup E(Q)tors of E(Q) is isomorphic to either Z/2Z ⊕ Z/2Z or
Z/2Z depending on whether n is square or not, respectively (cf. [9]). On the contrary, it is not so easy to determine the structure of the free part of
E(Q).
In [6] we investigated ranks of E(Q) and integer points on E for n =
pk with p prime and k ∈ {1, 2, 3}. While in some cases of rank one we determined the generators for E(Q) (e.g., in case k = 1 and p = (2t)2+ 1 with odd t, E(Q) = h(0, 0), (−1, 2t)i), we were not able to do that in rank two cases (e.g., in case k = 1 and p = a4+ b4> 17, the independence of the
points (−b2, a2b) and (−a2, ab2) was only found). Duquesne ([5, Theorem
12.3]) remarkably showed that if n = (2k2−2k+1)(18k2+30k+17) is
square-free with an integer k, then the points G1 = (−(2k2−2k +1), 4(k +1)(2k2− Manuscrit reçu le 11 mai 2010.
2k + 1)) and G2 = (9(2k2− 2k + 1), 12(3k − 2)(2k2− 2k + 1)) can always
be in a system of generators for E(Q) (where G1 and G2 above correspond
to G2 and G1+ G2 in [5], respectively. Note that he also determined the integer points on a quartic form of E assuming rankE(Q) = 2). His main strategy is to bound the canonical height ˆh on E(Q) in two ways. More precisely, he gave a uniform lower bound for the Archimedean part ˆλ∞ of
ˆ
h using Cohen’s algorithm ([4, Algorithm 7.5.7]), and gave upper bounds
for ˆλ∞(G1) and ˆλ∞(G2) using Tate’s series (cf. [13]). By combining them
with bounds for the non-Archimedean part ˆhfinof ˆh, he checked that in case
rankE(Q) = 2 Siksek’s theorem ([11, Theorem 3.1]) implies ν < 3, where ν denotes the lattice index of the span of G1 and G2 in E(Q)/E(Q)tors. Since
one easily see that ν 6= 2, this shows the result.
The major reasons Duquesne’s family worked well are that the x-coordi-nate x(Gi) of Gi (i ∈ {1, 2}) more or less divides n, which makes ˆhfin(Gi)
less than about −(log |x(Gi)|)/2, and that x(Gi) is similar in size to √n,
which keeps ˆλ∞(Gi) no larger than about (log n)/2, and hence, ˆh(Gi) =
ˆ
hfin(Gi) + ˆλ∞(Gi) is less than about (log n)/4 (see Lemma 3.2, Proposition
3.4 and its proof in Section 3). Moreover, putting s = 2k2 − 2k + 1 and
t = 18k2+30k +17 (then n = st), we found that G
1comes from the relation
t − s = and G2comes from the relation 81s − t =. These considerations
lead us to the following.
Theorem 1.1. Let n be a positive, non-square, fourth-power-free integer
such that n = st with positive, non-square integers s and t. Suppose that there exist positive integers α, β and m such that
t − s = α2 and m4s − t = β2.
(1.1)
Let E be the elliptic curve defined by
E : y2= x3− nx.
Then, the points G1 = (−s, sα) and G2 = (m2s, msβ) can always be in a
system of generators for E(Q) if m = 2 or 3. In case m ≥ 4, the same is true for n ≥ m26.
This paper is organized as follows. In Section 2, using the 2-descent lemma we show that the points G1 and G2 are independent. In Section 3,
following Duquesne’s strategy we estimate the canonical heights on E. In Section 4, applying Siksek’s theorem to the height estimates we complete the proof of Theorem 1.1. Finally in Section 5, we show that for each integer
m ≥ 2 there exist infinitely many binary forms n = n(k, l) in Z[k, l] each
of which represents infinitely many integers satisfying the assumptions in Theorem 1.1 (cf. Proposition 5.1 and the subsequent Remarks (1)), and give several examples (cf. Example just after the proof of Proposition 5.1)
of infinite families n = n(k, l), one of which contains Duquesne’s family (cf. Remark at the end of Section 5).
We now fix the notation. Let E be an elliptic curve defined by y2 =
x3− nx with a positive integer n. For this model of E, x(P ) denotes the
x-coordinate of a point P on E. For P = (x, y) in E(Q) with x = b/a
and gcd(a, b) = 1, the naïve height h : E(Q) → R is defined by h(P ) = log max{|a|, |b|}. The canonical height ˆh : E(Q) → R is defined by
ˆ h(P ) = lim n→∞ 1 4nh(2 nP )
(note that this value is double the definitions in [12, 13, 4]). The canonical height has a decomposition into local heights:
ˆ
h(P ) = X
p:prime or ∞
ˆ
λp(P ) for P ∈ E(Q) \ {O}.
We normalize the symbols ˆλpfollowing Duquesne’s paper (which are double
the definitions in [4], and satisfy ˆλp = 2(ˆλ0p+ log |∆|p/12), where ˆλ0p denote
the local heights defined in [13, 14]). Finally, for a prime number p denote by vp the valuation on Q normalized by vp(p) = 1.
2. Independence of the points G1 and G2
Let n, s, t, m, α, β be integers as in Theorem 1.1. Then, E(Q)tors= hT i '
Z/2Z, where T = (0, 0) (cf. [9]). In addition, E has the following Q-rational points
G1= (−s, sα), G2= (m2s, msβ).
Lemma 2.1. Let denote by n0the square-free part of n. On the assumptions in Theorem 1.1, any prime divisor p of n0 does not divide m2s − t.
Proof. Suppose that a prime divisor p of n0 divides m2s − t. Since p is
a divisor of n = st, it divides both ms and t. Then, m4s − t = β2 ≡ 0 (mod p2). Since n is fourth-power-free, we have either
vp(m4s) = vp(t) = 1
or
vp(m4s) ≥ 2 and vp(t) = 2.
Since p divides n0, the latter holds and vp(s) = 1, vp(t) = 2. This implies that vp(α2) = vp(t − s) = 1, which is a contradiction. Proposition 2.2. On the assumptions in Theorem 1.1, G1, G2, G1 + T,
G2+ T, G1+ G2, G1+ G2+ T 6∈ 2E(Q). Thus, G1 and G2 are independent
Proof. By the 2-descent lemma (cf. [9, Theorem 4.2]), if a point P = (x, y)
on E is in 2E(Q), then x, x +√n, x −√n must be squares in Q(√n). We
now have G1+ T = (t, tα), G2+ T = − t m2, tβ m3 , x(G1+ G2) = − mα + β m2+ 1 2 , x(G1+ G2+ T ) = n mα − β m2s − t 2
(note that s(m4 − 1) = α2+ β2 and (m2s + t)(m2− 1) = m2α2+ β2 by
assumption (1.1)). Hence, it is clear that G1, G2+ T, G1+ G2 6∈ 2E(Q).
Moreover, since s, t and n = st are non-square, the square-free part of n equals that of neither s nor t. Thus, s, t 6∈ Q(√n)2, that is, G2, G1+ T 6∈
2E(Q).
Suppose that G1 + G2 + T ∈ 2E(Q). Let n = n20n0 with square-free
integers n0, n0. Then, x(G1+ G2+ T ) + √ n ∈ Q(√n0)2 implies that (mα − β)2n + (m2s − t)2√n = (A + B √ n0)2
for some A, B with A, B ∈ Z or A, B ∈ 12Z \ Z. This means that (mα − β)2n = A2+ B2n0,
(2.1)
(m2s − t)2n0 = 2AB.
(2.2)
Clearly A, B ∈ Z. By (2.1) n0 divides A, and by (2.2) and Lemma 2.1 n0 divides n0. Hence by (2.1) n0 divides B, which contradicts (2.2) and Lemma
2.1. Therefore, G1+ G2+ T 6∈ 2E(Q).
Remarks. (1) Modifying the second equation of (1.1) a little, one can obtain an analogous result to the above proposition. More precisely, let
n = st and E be as in the proposition. Suppose that there exist positive
integers α, β and m with m even such that
t − s = α2 and t − m4s = β2.
Let G1 = (−s, sα) and G2 = (−m2s, msβ). Then, G1 and G2 are
inde-pendent modulo E(Q)tors. The reason this family did not work well is that
x(G1), x(G2) are not necessarily similar in size to
√
n. One can easily see
that the same is true for the family of a negative n = −st satisfying
t + s = α2 and t + m4s = β2
by replacing s with −s in the above argument.
(2) In the case where n is square, G1and G2 are not necessarily
indepen-dent modulo E(Q)tors. Indeed, let s = m2− 1 and t = m2(m2− 1) with a
positive integer m. Then, t−s = (m2−1)2, m4s−t = m2(m2−1)2, and both
s and t are non-square. On the other hand, since G1 = (1 − m2, (m2− 1)2)
and G2 = (m2(m2− 1), m2(m2− 1)2), we obtain G
1+ T = G2. Note that
3. Computation of the canonical heights on E
We begin by a brief summary of arithmetical properties of E. The dis-criminant of E is 64n3. Let p be a prime dividing n. By Tate’s algorithm ([15]), the reduction of E at p is of Kodaira type III, IM∗ or III∗ if re-spectively vp(n) = 1, 2 or 3, where M = 0 for odd p. The exponent of the conductor of E at p is 2 or 8 − M if respectively p is odd or p = 2, where if vp(n) = 2, then M corresponds to the subscript of IM∗ ; otherwise M = 0. If n is fourth-power-free and n 6≡ 0 (mod 4), then the sign ω(E) of the functional equation of the Hasse-Weil L-function L(E, s) is given by
ω(E) = −(n) · Y p2||n −1 p , (3.1)
where the product runs over odd primes, and
(n) =
(
−1 if n ≡ 1, 3, 11, 13 (mod 16), 1 otherwise
(see [1]). Periods of E can be expressed as follows.
Lemma 3.1. Let ω1, ω2 be periods of E such that ω1 > 0 and iω2 < 0.
Then, ω2= iω1 and ω1 ≥ π/(√2n14).
Proof. This lemma can be shown in the same way as Lemma 8.2 in [5].
Hereinafter, we give an uniform lower bound for the canonical height on
E and lower bounds for the canonical heights of the points G1 and G2.
The computation method follows Duquesne’s paper. The non-Archimedean part ˆhfin(P ) of ˆh(P ) can be computed by using Silverman’s algorithm ([13,
Theorem 5.2]). The computation of the Archimedean part ˆλ∞(P ) is crucial.
We compute ˆλ∞(P ) for any point P ∈ E(Q) \ E(Q)tors using the following
formula due to Cohen ([4, Algorithm 7.5.7]): ˆ λ∞(P ) = 1 16log 64n3 q +1 4log ω 1 2πy(P ) 2−1 2log |θ|, (3.2) where
q = e2πiω2ω1, ω1> 0, Im(ω2) > 0, Re(ω2) = 0,
either Im(z) = 0, 0 ≤ z < ω1 or Im(z) = Im(ω2/2), 0 ≤ z − ω2/2 < ω1,
and θ = ∞ X k=0 (−1)kqk(k+1)/2sin (2k + 1)2π ω1 Re(z)
with the elliptic logarithm z = z(P ) of P . Note that θ has a trivial bound |θ| < 1/(1 − |q|).
On the other hand, we compute ˆλ∞(G1) and ˆλ∞(G2) using Tate’s series (cf. [13]): ˆ λ∞(P ) = log |x(P )| + 1 4 N −1 X k=0 ck 4k + R(N ), (3.3) where ck= log |Z(2kP )|, Z(Q) = 1 + n x(Q)2 2 for Q ∈ E(Q) \ {(0, 0)}, 1 3 · 4N log (64n3)2 260H8 ! ≤ R(N ) ≤ 1 3 · 4N log 211H, (3.4) with H = max{4n, n2}.
We first compute the finite part ˆhfin(P ).
Lemma 3.2. For any point P = (a/d2, b/d3) in E(Q) with a, b, d ∈ Z, gcd(a, d) = gcd(b, d) = 1 and d > 0, we have
ˆ hfin(P ) = 2 log d − 1 2log Y pi|gcd(a,b,n), pi6=2 pei i + ˆh2(P ), (3.5) where pei
i kn with ei∈ {1, 2, 3}, and ˆh2(P ) is given by the following:
• If d is even, then ˆh2(P ) = 0.
• If d is odd, then the following holds.
n a b ˆh2(P )
even odd odd 0
odd even even 0
odd odd even −1
2log 2
v2(n) = 1 even even −12log 2
v2(n) = 2 and n/4 ≡ 1 (mod 4) v2(a) = 1 v2(b) ≥ 3 −32log 2
v2(n) = 2 and n/4 ≡ 3 (mod 4) v2(a) = 1 v2(b) = 2 −74log 2
v2(n) = 2 v2(a) ≥ 2 v2(b) ≥ 2 − log 2
v2(n) = 3 v2(a) ≥ 3 v2(b) ≥ 3 −32log 2
Proof. Since n is fourth-power-free, the equation y2 = x3 − nx is global minimal for E, and we may use Silverman’s algorithm in [13]. As mentioned at the beginning of this section, the reduction type of E at an odd prime p is III, I0∗ or III∗ if respectively vp(n) = 1, 2 or 3. Hence
ˆ
λp(P ) = −
1
where ψ3= 3a2(a2−2nd4)−n2d8. Noting that b2 = a(a2−nd4), one can see
that if p divides a, then vp(ψ3)/4 = vp(n)/2, and ˆλp(P ) = −(vp(P ) log p)/2.
The case p = 2 follows also from the algorithm through a case-by-case
argument.
We now bound ˆh(P ) below for any point P .
Proposition 3.3. Let n be a positive, fourth-power-free integer and E the
elliptic curve given by y2= x3− nx. If n 6≡ 12 (mod 16), then ˆ
h(P ) > 0.125 log n + 0.3917 for any non-torsion point P in E(Q).
Proof. Let P = (a/d2, b/d3) with a, b, d ∈ Z, gcd(a, d) = gcd(b, d) = 1 and
d > 0. By formula (3.2) and Lemma 3.1,
ˆ λ∞(P ) ≥ 1 16log 64n3 e−2π ! +1 4log ω1b2 2πd6 ! −1 2log 1 1 − e−2π ≥ π 8 − 1 2log 1 1 − e−2π + 1 8log n + 1 2log b d3 > 0.3917 + 1 8log n + 1 2log b d3 .
Combining this inequality with Lemma 3.2, we have ˆ h(P ) > 2 log d − 1 2log Y pi|gcd(a,b,n), pi6=2 pei i + ˆh2(P ) + 0.3917 +1 8log n + 1 2log b d3 = 0.125 log n + 0.3917 + 1 2log |bd| Y pi|gcd(a,b,n), pi6=2 pei i + ˆh2(P ).
Here, b2 = a(a2 − nd4) ensures that if p
i | a and peii | n, then p ei
i | b, and
the table in Lemma 3.2 implies that if n 6≡ 12 (mod 16), then v2(b) log 2 +
2ˆh2(P ) ≥ 0. Therefore we have 1 2log |bd| Y pi|gcd(a,b,n), pi6=2 pei i + ˆh2(P ) > 0,
and obtain the desired inequality.
Remarks. (1) Assumption (1.1) immediately implies that n = st 6≡ 12 (mod 16). Thus, we can use Proposition 3.3 in the proof of Theorem 1.1.
(2) Finding lower bounds for the canonical height of elliptic curves has been an active area of research. As for our curve E (with n not neces-sarily positive), Krir showed for any non-torsion point P in E(Q) that ˆ
h(P ) > (log |n|)/64 if n is fourth-power-free ([10, Proposition 4.1]), and
that ˆh(P ) > (log |n|)/16 if n is square-free ([10, Remarque 4.2]). Although
we are assuming n > 0 and n 6≡ 12 (mod 16), Proposition 3.3 gives a better bound than Krir’s ones.
We use Tate’s series to bound ˆλ∞(G1) and ˆλ∞(G2) above.
Proposition 3.4. On the assumptions in Theorem 1.1, ˆ h(G1) < 24577 98304log n + log m + 131081 196608log 2, ˆ h(G2) < 24577 98304log n + 1 2log m2m4+ 1+ 32777 196608log 2.
Proof. Since the discriminant 64n3 is positive, we have x(Q) ≥ √n for Q ∈ E0(R), the identity component of E(R). Hence, 1 ≤ Z(2kP ) ≤ 4 for P ∈ E(Q) and a positive integer k. It follows from (3.3) and (3.4) that for n ≥ 4, ˆ λ∞(P ) = log |x(P )| + 1 4 7 X k=0 ck 4k + R(8),
where c0 = 2 log(x(P )2+ n) − 4 log |x(P )|, 0 ≤ ci ≤ log 4 (1 ≤ i ≤ 7) and
R(8) ≤ (11 log 2 + 2 log n)/(3 · 48). Since G
1 = (−s, sα), G2 = (m2s, msβ) and s2+ n < 2n, m4s2+ n < (m4+ 1)n, we have ˆ λ∞(G1) < 49153 98304log n + 131081 196608log 2, ˆ λ∞(G2) < 49153 98304log n + 1 2log m4+ 1+ 32777 196608log 2. On the other hand, Lemma 3.2 together with√n < m2s implies that
ˆ hfin(Gi) ≤ − 1 2log s < − 1 4log n + log m
for i ∈ {1, 2}. Therefore, we obtain the desired inequalities. 4. Proof of Theorem 1.1
In order to prove Theorem 1.1, we need the following theorem of Siksek, based on the theory of quadratic forms (cf. [3]).
Theorem 4.1 (cf. [11, Theorem 3.1]). Let E be an elliptic curve over Q of rank r ≥ 2. Let G1 and G2 be independent points in E(Q) modulo
E(Q)tors. Choose a basis {P1, P2, . . . , Pr} for E(Q) modulo E(Q)tors such
order with ˆh(Q) ≤ λ, where λ is some positive real number. Then, the index ν of the span of G1 and G2 in hP1i + hP2i satisfies
ν ≤ √2 3 · p R(G1, G2) λ , where R(G1, G2) = ˆh(G1)ˆh(G2) − 1 4 ˆ h(G1+ G2) − ˆh(G1) − ˆh(G2) 2 . Proof of Theorem 1.1. In view of Proposition 2.2, in order to prove that G1
and G2can be in a system of the generators for E(Q), it suffices to show that
the lattice index ν is less than 3. By Proposition 3.3 and the subsequent Remarks (1), we may take λ = 0.125 log n + 0.3917. Since R(G1, G2) <
ˆ
h(G1)ˆh(G2), Proposition 3.4 and Theorem 4.1 together imply that ν <
f (m, n), where f (m, n) = √2 3· √ h1h2 0.125 log n + 0.3917 with h1= 24577 98304log n + log m + 131081 196608log 2, h2= 24577 98304log n + 1 2log(m 6+ m2) + 32777 196608log 2.
One can see that for a fixed m the function f (m, n) is decreasing. In the case of m = 2, if n ≥ 4885, then ν < f (2, n) < 3. The pairs (s, t) satisfying
n ≤ 4884 and the conditions in Theorem 1.1 are
(s, t) = (3, 39), (6, 15), (6, 87), (15, 159), (30, 39), (51, 87).
In each case, it is easy to check (e.g., by Magma ([2])) that G1and G2can be in a system of generators for E(Q). In fact, in the case of (s, t) = (15, 159),
E(Q) = h(0, 0), G1, G2, (−36, 198)i, and in all other cases, E(Q) = h(0, 0),
G1, G2i.
In the case of m = 3, if n ≥ 1.587 · 108, then ν < f (3, n) < 3. The number of those pairs (s, t) satisfying n < 1.587 · 108 and the assumptions in Theorem 1.1 is 2493. It is hard to check that G1 and G2 can be in a
system of generators for E(Q) directly. However, since f (3, 10) < 5, we have f (3, n) < 5 for all n ≥ 10. Since n ≤ 9 is not the case, it follows that
ν < 5. On the other hand, Proposition 2.2 implies that ν 6= 2, 4. Hence,
it suffices to show that ν 6= 3 for the 2493 pairs (s, t). We confirmed it by checking that none of G1, G2, G1+ G2 and G1− G2 has a three division
point in E(Q) for each (s, t) using the function “DivisionPoints(∗, 3)” in Magma ([2]).
In the case of m ≥ 4, consider the function g(m) = f (m, m26). Since g(m) is increasing and lim m→∞g(m) = 2 √ 3· r 24577 98304· 26 + 1 24577 98304· 26 + 3 0.125 · 26 < 2.9992 < 3, we have ν < 3 for n ≥ m26. This completes the proof of Theorem 1.1. Remark. There is no reason why the assertion in Theorem 1.1 does not hold for m ≥ 4 and n < m26. For example, one can easily see that G1 and
G2 can always be in a system of generators for 4 ≤ m ≤ 10 and s ≤ 30.
Indeed, since f (10, n) < 7 for n ≥ 66 and there is no n with n ≤ 65 satisfying the assumptions in Theorem 1.1, it suffices to check that
G1, G2, G1+ G2, G1− G26∈ 3E(Q),
G1, G2, G1+ G2, G1− G2, G1+ 2G2, G1− 2G2, 2G1+ G2, 2G1− G2 6∈ 5E(Q),
which can be done by Magma ([2]).
5. Construction of infinite families
By eliminating t from assumption (1.1), we have (m4− 1)s = α2+ β2.
Putting α = uk + vl and β = ul − vk yields
s = α 2+ β2 m4− 1 = u2+ v2 m4− 1(k 2+ l2) and t = s + (uk + vl)2.
Hence, u and v satisfying u2+ v2 ≡ 0 (mod (m4− 1)) give a binary form
n = st in Z[k, l]. This argument leads us to the following.
Proposition 5.1. Fix an integer m greater than one and write m4− 1 =
m0m1m22, where m0, m1, m2 are positive integers such that m0m1 is
square-free and any prime divisor of m0 (resp. m1) is congruent to 3 (resp. 1 or 2) modulo 4. Let p1, . . . , pr be distinct primes congruent to 1 or 2 modulo
4 such that none of the odd pi’s divides m4− 1 (possibly r = 0. If m1 = 1,
assume r ≥ 1; if m1 = p1 = 2, assume r ≥ 2). Let u0 and v0 be positive integers satisfying
(u0)2+ (v0)2 = m1p1· · · pr
(5.1)
and put u = m0m2u0 and v = m0m2v0. Then, the binary form n = st with
s = u
2+ v2
m4− 1(k
2+ l2) and t = s + (uk + vl)2
represents infinitely many integers satisfying the assumptions in Theorem
1.1.
Remarks. (1) Since there exist infinitely many primes congruent to 1 mod-ulo 4, Proposition 5.1 shows that for each integer m ≥ 2 there exist infinitely
many binary forms n = n(k, l) in Z[k, l] each of which represents infinitely many integers satisfying the assumptions in Theorem 1.1.
(2) A theorem of Granville ([8, Theorem 1]) implies that if ABC conjec-ture is valid, then each expression of n = st for each m ≥ 2 in Proposition 5.1 satisfies the assumptions in Theorem 1.1 for infinitely many integers k and infinitely many integers l.
Proof. We prove this proposition by applying to n/m20 or n/(2m20) a theo-rem of Gouvêa and Mazur ([7, Theotheo-rem 3]), which implies that if f (k, l) ∈ Z[k, l] is a binary form with nonzero discriminant, having no irreducible factor of degree exceeding three, and if the greatest common divisor of all values f (k, l) is free, then f (k, l) represents infinitely many square-free integers.
The direct computation shows that the discriminant of n is 16m4(u2 +
v2)12/(m4− 1)8, which is nonzero. Moreover, since
s = m0p1· · · pr(k2+ l2), t = s + (uk + vl)2
(5.2)
and u ≡ v ≡ 0 (mod m0), we always have s/m0, t/m0 ∈ Z[k, l], and if
m is odd and pi = 2 for some i, then both u and v must be even and
t/(2m0) ∈ Z[k, l]. Thus, it suffices to show that if m is odd and pi = 2 for
some i, then gcd(n; k, l ∈ Z) divides 2m20p1· · · pr; otherwise gcd(n; k, l ∈ Z)
divides m20p1· · · pr.
Expressing n and t as n(k, l) and t(k, l), respectively, we see that gcd(n;
k, l ∈ Z) divides
gcd(n(1, 0), n(0, 1), n(1, 1), n(1, −1)) = m0p1· · · prt0,
where t0= gcd(t(1, 0), t(0, 1), 2t(1, 1), 2t(1, −1)). If m is even, then u2+v2 =
m0(m4− 1)p1· · · pr implies that either u or v must be odd. Since
t = m 4u2+ v2 m4− 1 k 2+ 2uvkl + u2+ m4v2 m4− 1 l 2,
either t(1, 0) or t(0, 1) must be odd. Hence, the 2-primary part t0(2) of t0 equals 1. If m is odd, then both u and v must be even, and (5.2) implies that if pi = 2 for some i, then t0(2)= 2; otherwise t0(2)= 1.
It remains to examine the odd part t0odd of t0. By (5.2) t0odd divides gcd(m0p1· · · pr+ u2,m0p1· · · pr+ v2, 2m0p1· · · pr+ (u + v)2, 2m0p1· · · pr+ (u − v)2)odd, which divides gcd(2m0p1· · · pr+ u2+ v2, u2− v2, uv)odd (5.3) = gcd(m0p1· · · pr(m0m1m22+ 2), m20m22((u0)2− (v0)2), m20m22u0v0)odd,
where gcd(∗)odddenotes the odd part of gcd(∗). By gcd(m1, p1· · · pr)odd= 1
and (5.1) we have gcd(u0v0, m1p1· · · pr)odd= 1, and by
(u0)2− (v0)22 =(u0)2+ (v0)22− 4(u0)2(v0)2 = m21p21· · · p2
r− 4(u0)2(v0)2
we have gcd (u0)2− (v0)2, u0v0
odd = 1. Since gcd(m0m2, p1· · · pr)odd = 1,
(5.3) divides
m0gcd(m0m1m22+ 2, m0m22)odd= m0,
from which it follows that t0odd divides m0. Since we have already seen that gcd(n; k, l ∈ Z) divides m0p1· · · prt0, this completes the proof of Proposition
5.1.
From Proposition 5.1 one can easily obtain several parameterizations of
E in Theorem 1.1.
Example. Let k, l be nonzero integers. For each of the integers s, t ex-pressed in terms of k, l as listed below, the points G1 = (−s, sα) and
G2 = (m2s, msβ) with α =
√
t − s, β = √m4s − t and m = 2 or 3 can
always be in a system of generators for E(Q) if s, t, n = st are non-square and n is fourth-power-free. (1) The m = 2 cases: (a) s = 3(k2+ l2), t = 3(4k2+ 12kl + 13l2); (b) s = 6(k2+ l2), t = 3(5k2+ 18kl + 29l2); (c) s = 15(k2+ l2), t = 3(32k2+ 72kl + 53l2). (2) The m = 3 cases: (a) s = k2+ l2, t = 17k2+ 64kl + 65l2; (b) s = 2(k2+ l2), t = 2(9k2+ 48kl + 73l2); (c) s = 5(k2+ l2), t = 149k2+ 384kl + 261l2. Here, we took (u, v) in Proposition 5.1 as follows:
(1) (a) (u, v) = (3, 6); (b) (u, v) = (3, 9); (c) (u, v) = (9, 12). (2) (a) (u, v) = (4, 8); (b) (u, v) = (4, 12); (c) (u, v) = (12, 16). Remark. Duquesne’s family with s = 2k2− 2k + 1, t = 18k2+ 30k + 17 is
contained in the family with (2) (a) in Example above, since s = (1 − k)2+
k2, t = 17(1 − k)2+ 64(1 − k)k + 65k2.
We conclude this paper with a consideration about the rank. Let r be the rank of E(Q), and assume the parity conjecture. Then (−1)r = ω(E) holds, where ω(E) is the sign of the functional equation of L(E, s). If n is fourth-power-free and n 6≡ 0 (mod 4), then formula (3.1) implies that
(−1)r= −(n) Y p2||n −1 p ,
where if n ≡ 1, 3, 11, 13 (mod 16), (n) = −1; otherwise, (n) = 1. Consider the case (1) (a) in Example above. Noting n ≡ 0 (mod 9) and−13 = −1, one can see that if k ≡ 2 (mod 4) and n/9 is square-free, then r is odd with
r ≥ 3. Similarly, in the case (2) (a), if kl ≡ 2 (mod 4) and n is square-free,
then r is odd with r ≥ 3. We checked for 1 ≤ k, l ≤ 30 by Magma ([2]) that (1) (a) with k ≡ 2 (mod 4) and n/9 square-free has 70 cases, out of which 65 cases satisfy r ≥ 3, and (2) (a) with kl ≡ 2 (mod 4) and n square-free has 107 cases, out of which 81 cases satisfy r ≥ 3. In either case, it seems difficult to find a third generic point without further parameterizing k, l by quadratic or quartic binary forms.
Acknowledgment. The authors express their sincere thanks to the ref-eree for many valuable suggestions, which, in particular, made the lower bound for n in the case of m ≥ 4 in Theorem 1.1 better.
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Yasutsugu Fujita
College of Industrial Technology Nihon University
2-11-1 Shin-ei, Narashino, Chiba 275–8576 Japan
E-mail: fujita.yasutsugu@nihon-u.ac.jp
Nobuhiro Terai
Division of General Education Ashikaga Institute of Technology 268-1 Omae, Ashikaga, Tochigi 326–8558 Japan
